内容 1.3 Formulating Abstractions with Higher-Order Procedures · 58
练习 自检推理
Exercise 1.36: Modify fixed-point so that
it prints the sequence of approximations it generates, using the newline
and display primitives shown in Exercise 1.22. Then find a
solution to x
x
=
1000 by finding a fixed point of x
↦
log
(
1000
)
/
log
(
x
). (Use Scheme’s primitive log
procedure, which computes natural logarithms.) Compare the number of steps
this takes with and without average damping. (Note that you cannot start
fixed-point with a guess of 1, as this would cause division by
log
(
1
)
=
0.)
练习 1.36：修改 fixed-point，使其用练习 1.22 中给出的 newline 和 display 基本过程打印它所产生的近似值序列。然后，通过求 x ↦ log(1000)/log(x) 的不动点来找出 xˣ = 1000 的一个解。（使用 Scheme 的基本过程 log，它计算自然对数。）比较使用平均阻尼和不使用平均阻尼时所需步骤数的差异。（注意，不能以 1 作为 fixed-point 的初始猜测，因为这会导致除以 log(1) = 0。）
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